On Oct 25, 2008, at 2:47 PM, Robin van Spaandonk wrote:
In reply to Edmund Storms's message of Sat, 25 Oct 2008 09:06:07
-0600:
Hi Ed,
Robin, my main point is that an electron leaving an atom cannot go to
infinity under the conditions Mills has in his reactor. At most, it
will go into some other energy level, such as the conduction band if
one exists in the material. This fact is not based on speculation,
assumptions, or theory. This is a simple fact of nature that is well
understood.
[snip]
When an atom/molecule is ionized, the electron *never* goes to
infinity, so in
that sense, *no* measured (by *anyone*) ionization energy is 100%
accurate.
While that is true, the assumption is that the electron goes to
infinity.
However due to the inverse square drop in electric field, the
electron doesn't
have to be removed very far from an atom before the difference
between that and
infinity is so small as to be trivial (a few microns is enough).
Such distances
are easily attained in a plasma. What happens to the electron after
that is
irrelevant to the process from which the electron originated.
Suppose an electron goes from a level that requires 20 eV if the
electron goes to infinity. Now suppose the electron actually went to a
conduction band at 3 eV relative to infinity. Would not 17 eV be
required for the process?
In contrast, you assume that the final energy of the electron does not
matter provided it moves far enough from the original atom before
finding another state. If the Mills energy is based on this
assumption, then the environment in which the catalyst is located is
important. I agree, very little ambiguity is created when the
material is in a gas, as is most of the Mills work. However, we are
now talking about a solid mixture. I suggest this situation creates
great ambiguity and must be acknowledged.
In addition, I can imagine a range of energy being available in such a
transition if I can arbitrarily choose a distance the electron has to
move from its stable state before the energy being used is identified.
If this is the nature of the process, what is the point of choosing
the ionization energy as a criteria for the hydrino process working?
Regards.
Ed
Regards,
Robin van Spaandonk <[EMAIL PROTECTED]>
